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Unformatted text preview: Solutions to Homework # 3 1. We are in the situation of the asymmetric random walk with absorbing barriers. When the bets are $1, the barriers are at x = 0 and x = n and the walk starts at x = k . The probability that the walk hits x = n before hitting x = 0 is, therefore, (1.1) 1 ( q/p ) k 1 ( q/p ) n . When the bets are $0 . 5, the barriers are at x = 0 and x = 2 n and the walk starts at x = 2 k . The probability that the walk hits 2 n before hitting x = 0 is, therefore, (1.2) 1 ( q/p ) 2 k 1 ( q/p ) 2 n . If we divide (1.2) by (1.1) we get (1.3) 1 ( q/p ) 2 k 1 ( q/p ) k · 1 ( q/p ) n 1 ( q/p ) 2 n = 1 + ( q/p ) k 1 + ( q/p ) n . Now, if p > 1 / 2 then q < 1 / 2 and q/p < 1. In this case, ( q/p ) k > ( q/p ) n and hence (1.3) is greater than 1. Therefore, (1.2) is bigger than (1.1) and hence playing $0 . 50 is a better option. If p < 1 / 2 then q > 1 / 2 and q/p > 1. In this case, ( q/p ) n > ( q/p ) k and hence (1.3) is less than 1. Therefore, (1.1) is bigger than (1.2) and hence playing $1 is a better option. 2. We count trajectories that start at (0 , 10), end up at (50 , 20) and do not go below the line x = 5. Equivalently (shifting down by 4), we count trajectories that start at (0 , 6), end up at (50 , 16) and do not go below the line x = 1, that is, do not intersect the xaxis. First, we wind the total number of such trajectories: we find the number up of steps up and the number down of steps down from the system of equations up + down = 50 and up down = 10, so up = 30 and down = 20. Hence the total number of such trajectories is ( 50 20 ) . By the reflection principle, the number of trajectories that start at (0....
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 Winter '12
 AlexanderBarvinok
 Probability, Highways in Croatia, Croatia, Riga, trajectories

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